Yes, that's true. This is probably the most simplest and the most trickiest puzzle and it is guaranteed to keep classroom of highschool students including their teachers busy for at least a day.

My sister (who is currently in high school) brought it to me - it's kinda funny because she said that one of her classmates came up with it and said he saw it on internet and that the puzzle is solvable. As outcome the whole class became obsessed with it, including
one teacher (who reportedly did not listen to one pupil's homework report, because she was busy with solving ).

Anyway, do you think you have what it takes to solve this puzzle?

If you do, then here are the rules:
You have 16 lines connected like this.

Your objective is to draw one continuous line (that means you cannot hold up your pencil and start drawing somewhere else), which will cross all lines once and only once. The line you draw
doesn't need to be straight, is allowed to cross itself any can begin or end anywhere and you
don't need to finnish where you started drawing it.
Sounds pretty simple? Well, it often happens that you think you've solved it, but what really happen is that you forgot to cross one line or you crossed a line twice.

Try it yourself or show it to other people and watch how they'll waste tones of paper and go nuts with it.

Whilst this sure looks like a graph-theory puzzle, since you can eliminate the four even-degree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odd-degree is unsolvable.

Whilst this sure looks like a graph-theory puzzle, since you can eliminate the four even-degree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odd-degree is unsolvable.

Spills the beans.

I'm also stuck somehow. I tried to build partial graphs and connect them somehow, but I'm always missing one line. I have always three endpoints!

Whilst this sure looks like a graph-theory puzzle, since you can eliminate the four even-degree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odd-degree is unsolvable.

Spills the beans.

I'm also stuck somehow. I tried to build partial graphs and connect them somehow, but I'm always missing one line. I have always three endpoints!

Well, you have to create new graph by assigning every face its own point on the new graph, including the outer face.
The edge between two new points exist if you can directly draw line between the two faces that are on the first graph and represent points we are now connecting. The new edges can be multiple if two adjacent faces share more than one common edge that separate
them - the mutiplicity is then determined by the number of shared edges of two cells. In other words, we get new graph of possible pencil movements. Now we look if we have eulerian path (we have to be careful with points attached to edges that are multiple)

I'll post tomorrow a picture explaining it (I can't do it on the computer I am using now).

Anyway, I would like some nonmathematician solve it. It's the fun that never ends ..well at least it was for my family members, who tried to beat me by 'conventional' methods.

The problem I have come across is that when one crosses two of the sides of a shape it will leave 2 or 3 sides left. To cross either 2 or 3 sides, one must enter the shape over one side and exit over the other. This leaves the issue that one must be able
to enter the shape to cross certain lines. One could enter the shape and cross the line at the same time, however this leaves one inside a rectangle from which there is not escape as all lines have been crossed. I think this is what makes this puzzle difficult.